How to prove this nice $10$th power identity for $x_1^6+x_2^6+x_3^6 =y_1^6+y_2^6+y_3^6$?

Ramanujan's 6-10-8 Identity turns out to depend on a special case of, $$u_1^k+u_2^k+u_3^k =v_1^k+v_2^k+v_3^k$$ simultaneously valid for $k=2,4$. I was investigating if the next system $k=2,6$, $$x_1^2+x_2^2+x_3^2 =y_1^2+y_2^2+y_3^2\\x_1^6+x_2^6+x_3^6 =y_1^6+y_2^6+y_3^6\tag1$$ would have something similar. I observed empirically that, $$\left(\sum_{i=1}^3\big(x_i^{10}-y_i^{10}\big)\right)\left(\sum_{i=1}^3\big(x_i^{4}-y_i^{4}\big)\right)^2=20\prod_{i=1}^3\prod_{j=1}^3\big(x_i^2-y_j^2\big)\tag2$$ Example: $$10^k+15^k+23^k = 3^k+19^k+22^k$$ yields, $$\small \text{LHS}= \big(10^{10} + 15^{10} + 23^{10} - 3^{10} - 19^{10} - 22^{10}\big)\big(10^4 + 15^4 + 23^4 - 3^4 - 19^4 - 22^4\big)^2$$ $$\small \text{RHS}=20(10^2 - 3^2)(10^2 - 19^2)(10^2 - 22^2)(15^2 - 3^2)(15^2 - 19^2)(15^2 - 22^2)(23^2 - 3^2)(23^2 - 19^2)(23^2 - 22^2)$$ $$\small\text{LHS}=\text{RHS}=37739520^2\times3830610$$ I've also tested it with more general parametric solutions and it works just fine.

Q: But how do we prove $(2)$ rigorously?

• why so many squares ? Commented Dec 11, 2016 at 14:34
• @mercio: I observed this while working with $x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k$ for $k=2,4$. I'm not sure if the version $k=1,3$ and signed $x_i, y_i$ will work, but fortunately $(1)$ has infinitely many solutions to play with. Commented Dec 11, 2016 at 14:38
• @Tito Piezas III Is this question still of interest? Commented Jan 27, 2017 at 15:00
• @OldPeter: Feel free to answer it still. :) Commented Jan 27, 2017 at 16:34